Marie shoots an arrow vertically upward at a sandhill crane. If the arrow travels to a maximum height of 99.5 meters, calculate how fast it was shot.
And, for how much time was Marie's arrow in the air?
v = Vi - 9.8 t
0 = Vi - 9.8 t
Vi = 9.8 t
99.5 = Vi t -4.9 t^2
99.5 = 9.8 t^2 - 4.9 t^2 = 4.9 t^2
t = 4.5 seconds to top
so
Vi = 9.8 * 4.5 = 44.2 m/s
You could also do this by conservation of energy
(1/2) m Vi^2 = m g h
so
Vi = sqrt (2*9.8*99.5)
= 44.2 m/s
To calculate how fast the arrow was shot, we need to determine its initial velocity.
When an object is shot vertically upward, its initial velocity would be the same magnitude as its final velocity when it reaches its maximum height before falling back down. At the maximum height, the final velocity becomes 0 before it starts descending due to gravity.
We can use the equation: v^2 = u^2 - 2as
Where:
v = final velocity (0 m/s at max height)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2, negative because it acts in the opposite direction of the arrow's motion)
s = displacement (99.5 m, the maximum height reached)
Plugging in the values into the equation, we have:
0 = u^2 - 2(-9.8)(99.5)
Simplifying the equation:
0 = u^2 + 1946.6
To solve for u, we need to isolate it by moving 1946.6 to the other side of the equation:
u^2 = -1946.6
Now, take the square root of both sides to find the initial velocity:
u = √(-1946.6)
Note: Since the final velocity is zero, we take the positive square root since velocity cannot be negative.
u ≈ 44.13 m/s
Therefore, the arrow was shot approximately at a speed of 44.13 m/s.
To find the time the arrow was in the air, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time.
At the maximum height, the final velocity is 0, so we have:
0 = 44.13 - 9.8t
Solving for t:
9.8t = 44.13
t = 44.13 / 9.8
t ≈ 4.50 seconds
Therefore, Marie's arrow was in the air for approximately 4.50 seconds.